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FOC控制之CLARK和PARK变换

FOC控制之CLARK和PARK变换

CLARK变换

将abc坐标系转换为αβ静止坐标系

转换公式

{Uα=Uα−cos⁡(π/3)Ub−cos⁡(π/3)UcUβ=sin⁡(π/3)Ub−sin⁡(π/3)Uc\\begin{dcases} U_{α}=U_{α}-\\cos(\\pi/3)U_{b}-\\cos(\\pi/3)U_{c} \\\\ U_{β}=\\sin(\\pi/3)U_{b}-\\sin(\\pi/3)U_{c} \\end{dcases}{Uα=Uαcos(π/3)Ubcos(π/3)UcUβ=sin(π/3)Ubsin(π/3)Uc

转换矩阵

∣UαUβ∣=K∣1−1/2−1/20(3/2−(3/2∣∣UaUbUc∣\\begin{vmatrix} U_{α} \\\\ U_{β} \\end{vmatrix}=K \\begin{vmatrix} 1 & -1/2 & -1/2 \\\\ 0 & \\sqrt{\\mathstrut 3}/2 & -\\sqrt{\\mathstrut 3}/2 \\end{vmatrix}\\begin{vmatrix} U_{a} \\\\ U_{b} \\\\ U_{c} \\end{vmatrix}UαUβ=K101/2(3/21/2(3/2UaUbUc

PARK变换

将αβ静止坐标系转换为dq旋转坐标系

转换公式

{Ud=Uαcos⁡(θ)+Uβsin⁡(θ)Uq=−Uαsin⁡(θ)+Uβcos⁡(θ)\\begin{dcases} U_{d} = U_{\\alpha}\\cos(\\theta) + U_{\\beta}\\sin(\\theta) \\\\ U_{q} = -U_{\\alpha}\\sin(\\theta) + U_{\\beta}\\cos(\\theta) \\end{dcases}{Ud=Uαcos(θ)+Uβsin(θ)Uq=Uαsin(θ)+Uβcos(θ)

转换矩阵

[UdUq]=[cos⁡(θ)sin⁡(θ)−sin⁡(θ)cos⁡(θ)][UαUβ]\\begin{bmatrix} U_{d} \\\\ U_{q} \\end{bmatrix}=\\begin{bmatrix} \\cos(\\theta) & \\sin(\\theta) \\\\ -\\sin(\\theta) & \\cos(\\theta) \\end{bmatrix}\\begin{bmatrix} U_{\\alpha} \\\\ U_{\\beta} \\end{bmatrix}[UdUq]=[cos(θ)sin(θ)sin(θ)cos(θ)][UαUβ]

PARK逆变换

变换公式

{Uα=Udcos⁡(θ)−Uqsin⁡(θ)Uβ=Udsin⁡(θ)+Uqcos⁡(θ)\\begin{dcases} U_{\\alpha} = U_{d}\\cos(\\theta) - U_{q}\\sin(\\theta) \\\\ U_{\\beta} = U_{d}\\sin(\\theta) + U_{q}\\cos(\\theta) \\end{dcases}{Uα=Udcos(θ)Uqsin(θ)Uβ=Udsin(θ)+Uqcos(θ)

变换矩阵

[UαUβ]=[cos⁡(θ)−sin⁡(θ)sin⁡(θ)cos⁡(θ)][UdUq]\\begin{bmatrix} U_{\\alpha} \\\\ U_{\\beta} \\end{bmatrix}=\\begin{bmatrix} \\cos(\\theta) & - \\sin(\\theta) \\\\ \\sin(\\theta) & \\cos(\\theta) \\end{bmatrix}\\begin{bmatrix} U_{d} \\\\ U_{q} \\end{bmatrix}[UαUβ]=[cos(θ)sin(θ)sin(θ)cos(θ)][UdUq]

CLARK逆变换

变换公式

{Ua=UαUb=cos⁡(2π/3)Uα+sin⁡(π/3)UβUc=cos⁡(2π/3)Uα−sin⁡(π/3)Uβ\\begin{dcases} U_{a} = U_{\\alpha} \\\\ U_{b} = \\cos(2\\pi/3)U_{\\alpha}+\\sin(\\pi/3)U_{\\beta} \\\\ U_{c} = \\cos(2\\pi/3)U_{\\alpha}-\\sin(\\pi/3)U_{\\beta} \\end{dcases}Ua=UαUb=cos(2π/3)Uα+sin(π/3)UβUc=cos(2π/3)Uαsin(π/3)Uβ

变换矩阵

∣UaUbUc∣=∣10−1/2(3/2−1/2−(3/2∣∣UαUβ∣\\begin{vmatrix} U_{a} \\\\ U_{b} \\\\ U_{c} \\end{vmatrix}=\\begin{vmatrix} 1 & 0 \\\\ -1/2 & \\sqrt{\\mathstrut 3}/2 \\\\ -1/2 & -\\sqrt{\\mathstrut 3}/2 \\end{vmatrix}\\begin{vmatrix} U_{\\alpha} \\\\ U_{\\beta} \\end{vmatrix}UaUbUc=11/21/20(3/2(3/2UαUβ